1. **State the problem:** We need to find the value of $x$ and several angle measures given that $AR \cong RO \cong SA$ and the angles $m\angle AMR = 3x + 20$ and $m\angle ONR = x + 30$. 2. **Use the equality of angles:** Since $AR \cong RO$, angles $AMR$ and $ONR$ are equal: $$3x + 20 = x + 30$$ 3. **Solve for $x$:** $$3x + 20 = x + 30$$ $$3x - \cancel{x} + 20 = \cancel{x} + 30$$ $$2x + 20 = 30$$ $$2x = 30 - 20$$ $$2x = 10$$ $$x = 5$$ 4. **Calculate the angles:** - $m\angle AMR = 3(5) + 20 = 15 + 20 = 35^\circ$ - $m\angle ONR = 5 + 30 = 35^\circ$ 5. **Find $m\angle ANR$:** Since $ONR$ and $ANR$ form a linear pair, $$m\angle ANR = 180^\circ - 35^\circ = 145^\circ$$ 6. **Other angles by symmetry and properties:** - $m\angle ORM = m\angle AMR = 35^\circ$ - $m\angle AM = 60^\circ$ (equilateral triangle property) - $m\angle RNO = 35^\circ$ - $m\angle RAM = 60^\circ$ - $m\angle AR = 35^\circ$ - $m\angle OM = 35^\circ$ - $m\angle ROM = 60^\circ$ - $m\angle AMO = 180^\circ - 35^\circ = 145^\circ$ **Final answers:** - $x = 5$ - $m\angle ANR = 145^\circ$ - $m\angle ORM = 35^\circ$ - $m\angle AM = 60^\circ$ - $m\angle RNO = 35^\circ$ - $m\angle RAM = 60^\circ$ - $m\angle AR = 35^\circ$ - $m\angle OM = 35^\circ$ - $m\angle ROM = 60^\circ$ - $m\angle AMO = 145^\circ$